November 8th
Today I learned that it's possible to do calculus over ratios of prime numbers, from here . The general theorem is that\[\lim_{n\to\infty}\frac1n\sum_{k=1}^nf\left(\frac{p_k}{p_n}\right)=\int_0^1f(x)\,dx\]where $\{p_\bullet\}$ is the sequence of primes. Here we are assuming that $f$ is Riemann integrable already, or else it doesn't make much sense to conjecture this. Anyways, it suffices to suppose that $f$ is continuously differentiable; if not we can certainly approximate $f$ well enough by continuously differentiable functions.
The way we translate the summation into an integral is to use Riemann-Stieltjes integration. Write\[\sum_{k=1}^nf\left(\frac{p_k}{p_n}\right)=\int_0^1f(x)\,d\pi(p_nx).\]Integrating by parts, this is\[\sum_{k=1}^nf\left(\frac{p_k}{p_n}\right)=nf(1)-\int_0^1f'(x)\pi(p_nx)\,dx.\]To finish getting rid of the prime stuff, we would like to substitute $\pi(x)\sim\frac x{\log x}$ from the prime number theorem, but we can't do that while the lower bound is $0.$ However, we may bound\[\left|\int_0^{1/\log n}f'(x)\pi(p_nx)\,dx\right|\le\left(\max_{x\in[0,1]}\left|f'(x)\right|\right)\pi\left(\frac{p_n}{\log n}\right).\]The maximum is a constant, so we may more or less ignore it. The prime-counting term can be estimated to $o(n),$ which is good enough because we're going to divide by $n$ later. Indeed, $\pi(n)\sim\frac n{\log n}$ implies $n\sim\frac{p_n}{\log p_n}.$ However, this means $\log n=\log p_n+\log\log p_n+o(1)=\log p_n+o(\log p_n),$ implying $p_n\sim n\log p_n=n\log n(1+o(1))\sim n\log n.$ Thus, $\frac{p_n}{\log n}\sim n,$ so $\pi\left(\frac{p_n}{\log n}\right)=o(n).$ I'm skimming some details because I don't care about them. It follows that this entire term is $o(n).$
So we are left with\[\sum_{k=0}^nnf\left(\frac{p_k}{p_n}\right)=nf(1)-\int_{1/\log n}^1f'(x)\pi(p_nx)\,dx+o(n).\]The prime number theorem says that $\pi(p_nx)\sim\frac{p_nx}{\log(p_nx)}.$ However, now that $x\in[1/\log n,1],$ the denominator is between $\log(p_n/\log n)$ and $\log(p_n).$ With $n\sim\frac{p_n}{\log p_n}$ implying $\log p_n=\log n+o(\log n),$ we get that both of these are asymptotically $\log n.$ Explicitly, we see\[\pi(p_nx)\sim\frac{p_nx}{\log n}\sim nx\]after again plugging in $p_n\sim n\log n.$ This means that the integral is\[\int_{1/\log n}^1f'(x)(nx+o(nx))\,dx=n\int_{1/\log n}^1xf'(x)\,dx+o\left(n\int_{1/\log n}^1xf'(x)\,dx\right).\]The bounding term is (again) lazily $o(n)$ because the full integral is constant. To actually evaluate the integral, we begin by re-extending the lower bound to $0.$ The remainder added amount from $0$ to $\frac1{\log n}$ is $o\left(n\int_0^{1/\log n}x\right)=o(n),$ so we are left with\[n\int_0^1xf'(x)\,dx+o(n).\]To finish, we integrate by parts in reverse, which makes this\[nf(1)-\int_0^1f(x)\,dx+o(n).\]
Putting all the pieces together, we get that\[\sum_{k=1}^nf\left(\frac{p_k}{p_n}\right)=nf(1)-nf(1)+\int_0^1f(x)\,dx+o(n).\]Dividing by $n$ and taking $n\to\infty$ gives the result.
It would be something of a crime to not do any applications of this result. We can asymptotically compute the arithmetic mean of the first $n$ primes. Namely, note that\[\lim_{n\to\infty}\frac{\displaystyle\frac1n\sum_{k=1}^np_k}{p_n}=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac{p_k}{p_n}=\int_0^1x\,dx=\frac12.\]It follows that the arithmetic mean is asymptotically $\frac12p_n=\frac12n\log n.$ We can even do the geometric mean. Note that\[\lim_{n\to\infty}\frac1n\sum_{k=1}^n\log\left(\frac{p_k}{p_n}\right)=\int_0^1\log x\,dx=-1.\]This implies that\[\lim_{n\to\infty}\frac{\sqrt[n]{p_1\cdots p_n}}{p_n}=\lim_{n\to\infty}\prod_{k=1}^n\frac{p_k}{p_n}=\exp\left(\lim_{n\to\infty}\frac1n\sum_{k=1}^n\log\left(\frac{p_k}{p_n}\right)\right)=\frac1e.\]It follows that the geometric mean is asymptotically $\frac1ep_n\sim\frac1en\log n.$ As expected, (the asymptotics for) the geometric mean is smaller than the arithmetic mean.